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Mariana [72]
3 years ago
7

Im a bit confused so please help me quick!

Mathematics
2 answers:
Ainat [17]3 years ago
4 0
Yes you are correct hope this helps!
SVETLANKA909090 [29]3 years ago
3 0

Answer:

yes you are correct

Step-by-step explanation:

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If 16 oz equal 1 pound how much ounces are equal to 3 pounds
Solnce55 [7]

Answer:

it would be 48oz

Step-by-step explanation:

16 times 3 = 48oz -sorry if this is not right

have a wonderful day :)

3 0
2 years ago
Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

4 0
3 years ago
Can someone please help me with this
Iteru [2.4K]
The answer is 5 3/4.............
8 0
2 years ago
Read 2 more answers
While hiking, Sanjay went up 200 meters. If Sanjay started at 500 meters above sea level, which integers represents his elevatio
Aliun [14]

Answer:300

Step-by-step explanation:

-200+500=300

3 0
3 years ago
What is -138 divided by 18
balu736 [363]
-138/18= -7.6 repeating
4 0
3 years ago
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